Answer:
12.6332454263 m/s
Explanation:
m = Mass of car
v = Velocity of the car
= Coefficient of static friction = 0.638
g = Acceleration due to gravity = 9.81 m/s²
r = Radius of turn = 25.5 m
When the car is on the verge of sliding we have the force equation
The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s
Answer:
-48 N
Explanation:
mass of door (m) = 4 kg
acceleration of the door = 12 m/s^{2}
force exerted by the person = 48 N
From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N
Use kinematic equations to solve:
1) yf = yo + vo*t + 1/2at²
yf = final height
yo = initial height
vo = initial velocity
a = acceleration
t = time
yf - yo = vo*t + 1/2at²
yf - yo = h
vo = 0
Thus,
h = 1/2at²
h = 1/2(9.8)(12)² = 705.6 m
2) vf = vo + at
vo = 0
Thus,
vf = at
vf = (9.8)(12) = 117.6 m/s
D is the best answer. In many physics problems we treat an extended object as if it were a point with the same mass located at the center of mass.
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